Hi, there could u guys help me to get this problem??? thx!!!
v(t)=t^2 - 2t, 0(
answer is 8/3 but i cannot get the right answer!! help me thx
2006-11-18 07:52:33 · 4 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
assuming 0 and 3 are limits of integration:
v(t) = dx/dt = t^2 - 2t
x = (1/3)t^3 - t^2 + C
x = (1/3)3^3 - 3^2 - 0 + 0
x = 9 - 9 = 0
If the limits are 0 and 2,
x = 8/3 - 4 = -4/3
Going backwards,
(1/3)t^3 - t^2 = 8/3
t^3 - 3t^2 -8 = 0
t ≈ 3.612887865
2006-11-18 08:35:46 · answer #1 · answered by Helmut 7 · 0⤊ 0⤋
Abe got the eq right, but multiplied t by 2 instead of squaring it....
The answer is therefore t^3/3 - t^2 = 27/3 - 9 = 0
The actual motion of this particle begins with a negative velocity which changes to + at t = 1 sec. At 3 sec the particle has arrived back where it started for a net displacement of zero.
2006-11-18 16:21:46 · answer #2 · answered by Steve 7 · 0⤊ 0⤋
distance travelled ,s =[ v(t)^2 - v(0)^2 ]/2a
where a = {v(t)-v(0)}/t
here t =3
v(0) =0
v(3) = 3
hence a = 1
therefore s = 9/2
should be the answer. or check any other details given in the question.
2006-11-18 16:02:27 · answer #3 · answered by anami 3 · 0⤊ 0⤋
s(t) = t^3/3 - t^2
s(3) = 9-6 = 3
Maybe you need to check how the problem is stated in the text and the answer back in the book.
Please check for errors on my part.
2006-11-18 16:00:41 · answer #4 · answered by Anonymous · 0⤊ 1⤋